Extract arguments from stdout and pipe

I was trying to execute a script n times with a different fi开发者_如何学Pythonle as argument each time using this:

ls /user/local/*.log | xargs script.pl

(script.pl accepts file name as argument)

But the script is executed only once. How to resolve this ? Am I not doing it correctly ?

 ls /user/local/*.log | xargs -rn1 script.pl

I guess your script only expects one parameter, you need to tell xargs about that.

Passing -r helps if the input list would be empty

Note that something like the following is, in general, better:

 find /user/local/ -maxdepth 1 -type f -name '*.log' -print0 |
       xargs -0rn1 script.pl

It will handle quoting and directories safer.

To see what xargs actually executes use the -t flag.

I hope this helps:

for i in $(ls /usr/local/*.log)
do
  script.pl $i
done

I would avoid using ls as this is often an alias on many systems. E.g. if your ls produces colored output, then this will not work: for i in ls; do ls $i;done

Rather, you can just let the shell expand the wildcard for you: for i in /usr/local/*.log; do script.pl $i; done